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Projectile motionPhysicsHigh schoolabout 1:26

A projectile is launched at 25 m/s at 60° from a height of 0 m. Find its time of flight, range and maximum height: animated lesson

A narrated, step-by-step projectile motion video lesson with captions. Final answer: Time of flight ≈ 4.4139929 s; range ≈ 55.174911 m; maximum height ≈ 23.891437 m; impact speed ≈ 25 m/s.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.6 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    A projectile is launched at 25 m/s at 60° from a height of 0 m. Find its time of flight, range and maximum height.
  2. 2. Steps, part 1
    1. vx=v0cos⁡θ=25cos⁡60∘=12.5,vy=v0sin⁡θ=21.650635 m/sv_x = v_0\cos\theta = 25\cos 60^\circ = 12.5,\quad v_y = v_0\sin\theta = 21.650635\ \text{m/s}
      Split the launch velocity into components
    2. tpeak=vyg=2.2069964 st_{\text{peak}} = \frac{v_y}{g} = 2.2069964\ \text{s}
      Time to the highest point
    3. H=h0+vy22g=23.891437 mH = h_0 + \frac{v_y^2}{2g} = 23.891437\ \text{m}
      Maximum height
    4. h0+vyt−12gt2=0⇒t=vy+vy2+2gh0g=4.4139929 sh_0 + v_y t - \tfrac12g t^2 = 0 \Rightarrow t = \frac{v_y + \sqrt{v_y^2 + 2gh_0}}{g} = 4.4139929\ \text{s}
      Time of flight (land at height 0)
  3. 3. Steps, part 2
    1. R=vxt=12.5×4.4139929=55.174911 mR = v_x t = 12.5 \times 4.4139929 = 55.174911\ \text{m}
      Horizontal range
    2. vy(T)=vy−gT=−21.650635,∣v∣=vx2+vy(T)2=25 m/s at 60∘ below horizontalv_y(T) = v_y - gT = -21.650635,\quad |v| = \sqrt{v_x^2 + v_y(T)^2} = 25\ \text{m/s at } 60^\circ \text{ below horizontal}
      Impact velocity
  4. 4. The answer
    T≈4.4139929 s, R≈55.174911 m, H≈23.891437 mT \approx 4.4139929\ \text{s},\ R \approx 55.174911\ \text{m},\ H \approx 23.891437\ \text{m}

    T = 4.4139929 s, R = 55.174911 m, H = 23.891437 m

  5. 5. Check
    • Energy check: v² at impact = 625 and v₀² + 2gh₀ = 625 (equal).
    • Level-ground formula: v₀² sin 2θ / g = 55.174911 m.

Transcript

The problem. Here is the problem: A projectile is launched at 25 m/s at 60° from a height of 0 m. Find its time of flight, range and maximum height. Treat horizontal and vertical motion separately: constant velocity horizontally, constant downward acceleration g vertically, ignoring air resistance.

Steps, part 1. Split the launch velocity into components. Time to the highest point. Maximum height. Time of flight (land at height 0).

Steps, part 2. Horizontal range. Impact velocity.

The answer. So the answer is Time of flight approximately 4.4139929 s; range approximately 55.174911 m; maximum height approximately 23.891437 m; impact speed approximately 25 m/s. Without air resistance the horizontal velocity never changes, while the vertical velocity falls by g every second. The flight ends when the height returns to zero.

Check. Energy check: v squared at impact equals 625 and v₀ squared plus 2gh₀ equals 625 (equal). Level-ground formula: v₀ squared sin 2theta / g equals 55.174911 m.

About projectile motion

Time of flight, range and maximum height of a projectile.

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